Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <queue>
#include <vector>
#include <map>
using namespace std;

map<int, vector<int> > edges;

// ?????¢?????????
int s2t[100001];
int t2s[100001];

// ???????????¢???????????°
int s2t_kinds[100001];
int t2s_kinds[100001];

void bfs(int s, int* kyori) {
    queue<pair<int, int> > q;
    q.push( make_pair(s,0) );

    while (!q.empty()) {
        pair<int,int> p = q.front(); q.pop();
        int curr = p.first;
        int step = p.second;
        for (int i=0; i<edges[curr].size(); i++) {
            int next = edges[curr][i];
            if (*(kyori + next) != -1) continue;
            *(kyori + next) = step + 1;
            q.push( make_pair(next, step + 1) );
        }
    }
}

int main() {
    int n,m,s,t; cin >> n >> m >> s >> t;
    for (int i=0; i<m; i++) {
        int x,y; cin >> x >> y;
        if (edges.count(x) == 0) {
            vector<int> v; v.push_back(y);
            edges[x] = v;
        } else {
            edges[x].push_back(y);
        }
        if (edges.count(y) == 0) {
            vector<int> v; v.push_back(x);
            edges[y] = v;
        } else {
            edges[y].push_back(x);
        }
    }

    // init
    for (int i=0; i<=n; i++) s2t[i] = t2s[i] = -1;
    s2t[s] = t2s[t] = 0;

    bfs(s,s2t);
    bfs(t,t2s);

    for (int i=1; i<=n; i++) {
        if (s2t[i] != -1) {
            s2t_kinds[ s2t[i] ] += 1;
        }
        if (t2s[i] != -1) {
            t2s_kinds[ t2s[i] ] += 1;
        }
    }

    int dist = s2t[t];
    long long ans = 0;
    for (int i=0; i<=dist-2; i++) {
        ans += (long long)s2t_kinds[i] * (long long)t2s_kinds[dist - 2 - i];
    }

    cout << ans << endl;


    return 0;
}